Volume generated by rotating around y-axis, curve $y=x^3$ and the lines $y=0$ and $x=2$

Question

Find the volume of the solid generated by revolving about the y-axis the region bounded by the curve $y=x^3$ and the lines $y=0$ and $x=2$

I first found what $x=2$ would be in terms of y.

$$y= (2)^3 \\ = 8$$

And in terms of y, the original equation becomes:

$$y=x^3 \\ x= y^{1 \over 3}$$

So,

$$V= \int^8_0 (y^{1 \over 3})^2 \pi dy \\ = \pi \int_0^8 y^{2\over 3}dy \\ =\pi \bigg[ y^{5\over 3} ({3 \over 5}) \bigg]^8_0 \\ = (8^{5 \over 3})({3 \over 5})\pi \\ = {96 \over 5} \pi$$

Therefore the answer is ${96 \over 5} \pi$ units $^3$

However, the answer is supposed to be ${64 \over 5} \pi$ units${^3}$

What went wrong?

Answer 1

The shell method would be easiest to use here:

$$ V=2\pi\int_{0}^{2}x\cdot x^3\,dx=2\pi\int_{0}^{2}x^4\,dx= 2\pi\frac{x^5}{5}\bigg|_{0}^{2}=\\ \frac{2\pi}{5}\left(2^5-0^5\right)= \frac{2\pi}{5}\cdot 32=\frac{64\pi}{5}\ \text{cubic units}. $$

Or you can use the washer method which is going to be a bit messier. In that case, you're subtracting the volume of a cylinder of radius $2$ and height $8$ from the volume you get by revolving the curve $x=\sqrt[3]{y}$ around the $y$-axis:

$$ V=\int_{0}^{8}\left(\pi\cdot 2^2-\pi\left(\sqrt[3]{y}\right)^2\right)\,dy= \pi\int_{0}^{8}\left(4-y^{\frac{2}{3}}\right)\,dy=\frac{64\pi}{5}\ \text{cubic units}. $$

Wolfram Alpha check.

Answer 2

The volume required is between the lines y=0, x=2 and curve y=x^3.

If you visualize it, the required volume is : V1-V2 where V1 = the volume of cylinder formed by line y=2 and x=2. V2 = the volume enclosed by the curve y=x^3 around y axis.

V1 = pi*r^2*h. r=2, h = 8.

so V1 = 4*8*pi = 32 pi V2 = 96/5 pi V1-V2 = 32pi - 96/5pi = 64/5 pi

Please pardon me as I dont know the mathML.

Answer 3

You are calculating the empty volume between the rotated function and the y-axis. This is because for every y-value, you are summing the distance between the y-axis and the function y=x^3, rather than the distance between the function y=x^3 and your boundary of x=2. As mentioned by someone in the comments, you want

$$V = \int_0^8 4-y^{2/3}dy$$

Answer 4

Washer method is quite easy. Express the washer area including its hole and extrude it along $y$.

$$ V=\pi\int_{0}^{8}\left(2^2-x^2\right)\,dy=\pi\int_{0}^{2}\left(2^2-x^2\right)\,3x^2 dx= $$ $$ V=3\pi\int_{0}^{2}\left(4-x^2\right)\, \cdot x^2 \, dx = ... = \frac{64 \pi}{5}$$